/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 11 A 50.0 -m length of coaxial cabl... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A 50.0 -m length of coaxial cable has an inner conductor that has a diameter of 2.58 \(\mathrm{mm}\) and carries a charge of \(8.10 \mu \mathrm{C} .\) The surrounding conductor has an inner diameter of 7.27 \(\mathrm{mm}\) and a charge of \(-8.10 \mu \mathrm{C} .\) (a) What is the capacitance of this cable? (b) What is the potential difference between the two conductors? Assume the region between the conductors is air.

Short Answer

Expert verified
The capacitance of the cable is given by the formula \(C = \frac{2\pi\epsilon_0 L}{\ln(b/a)}\). After calculation, the potential difference \(V\) is found using \(V = \frac{Q}{C}\).

Step by step solution

01

- Convert Units

Before calculating the capacitance, we need to ensure all units are consistent. We have diameters in millimeters, charge in microcoulombs, and length in meters. Convert the diameters to meters, and the charge to coulombs. For diameter of the inner conductor, this is 2.58 mm = 0.00258 m. For the diameter of surrounding conductor, 7.27 mm = 0.00727 m. The charge, 8.10 µC = 8.10 × 10^{-6} C.
02

- Calculate Capacitance

The capacitance of the coaxial cable can be found using the formula for the capacitance of a cylindrical capacitor: \(C = \frac{2\pi\epsilon_0 L}{\ln(b/a)}\). Here, \(\epsilon_0\) is the vacuum permittivity (\(8.85 \times 10^{-12} F/m\)), \(L\) is the length of the cable, \(a\) is the radius of the inner conductor, and \(b\) is the inner radius of the surrounding conductor. First, we need to calculate the radii: \(a = \frac{0.00258 m}{2}\) and \(b = \frac{0.00727 m}{2}\). Then, plug in all values into the capacitance formula to find the capacitance.
03

- Solve for Potential Difference

To find the potential difference (\(V\)) between the conductors, use the relation \(V = \frac{Q}{C}\), where \(Q\) is the charge on one of the conductors (since they have equal and opposite charges, we can use the magnitude of one charge) and \(C\) is the capacitance calculated in Step 2. Solve this equation to find the potential difference.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Cylindrical Capacitor
A cylindrical capacitor, as found in coaxial cables, consists of two concentric cylindrical conductors, usually separated by a dielectric material. In a coaxial cable, the core conductor is encased within an insulating layer, upon which the second conductor (or shield) is wrapped, creating a cylinder around the core. The capacitance of this system is determined by the physical properties of these conductors and the insulating material between them.

The efficient design of coaxial cables minimizes the leakage of electrical energy and maintains the separation of different signal channels, which makes them widely used in the transmission of high-frequency electrical signals over long distances.
Capacitance Calculation
To calculate the capacitance of a cylindrical capacitor, one can use the formula:
\[C = \frac{2\pi\epsilon_0 L}{\ln(b/a)}\]
where \(C\) represents the capacitance, \(\epsilon_0\) is the vacuum permittivity, \(L\) is the length of the coaxial cable, \(b\) is the inner radius of the outer conductor, and \(a\) is the radius of the inner conductor. The natural logarithm of the ratio of radii \(\ln(b/a)\) appears in the equation due to the geometry of the cylindrical structure. It's essential to use consistent units, converting all measurements to meters and farads for accuracy in capacitance calculations.
Potential Difference
The potential difference, often referred to as voltage, between the two conductors in a coaxial cable is an important aspect as it determines the electric field strength in the cable. The potential difference can be calculated once the capacitance is known, using the formula:
\[V = \frac{Q}{C}\]
where \(V\) is the potential difference, \(Q\) is the charge on one of the conductors, and \(C\) is the capacitance. A higher potential difference could lead to a stronger electric field, which may affect the performance of the coaxial cable, hence understanding this concept is crucial for troubleshooting and optimizing cable design.
Vacuum Permittivity
Vacuum permittivity, symbolized as \(\epsilon_0\), is a physical constant that describes the ability of a vacuum to permit electric field lines. Its value is approximately \(8.85 \times 10^{-12} \text{F/m}\) (farads per meter). This constant is crucial in calculations involving electric fields and capacitance, especially in the design of capacitors where it acts as a scaling factor. It forms part of the fundamental equations used for calculating the capacitance of cylindrical capacitors and reflects the electric permittivity of free space, acting as the reference point for other materials.
Charge Conversion
Charge conversion is necessary when dealing with electrical quantities, as it ensures that calculations are performed with consistent units. In the context of the coaxial cable problem, the charge is initially given in microcoulombs (\(\mu C\)), where one microcoulomb equals \(1 \times 10^{-6}\) coulombs. Converting the charge to coulombs before using it in capacitance and potential difference calculations is crucial to obtaining accurate results. This step prevents errors and ensures that the formulas yield values in standard units, which can then be properly interpreted and applied.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A small rigid object carries positive and negative 3.50 -nC charges. It is oriented so that the positive charge has coordinates \((-1.20 \mathrm{mm}, 1.10 \mathrm{mm})\) and the negative charge is at the point \((1.40 \mathrm{mm},-1.30 \mathrm{mm} \text { ). (a) Find the electric }\) dipole moment of the object. The object is placed in an electric field \(\mathbf{E}=(7800 \hat{\mathbf{i}}-4900 \hat{\mathbf{j}}) \mathrm{N} / \mathrm{C}\) . (b) Find the torque acting on the object. (C) Find the potential energy of the object- field system when the object is in this orientation. (d) If the orientation of the object can change, find the difference between the maximum and minimum potential energies of the system.

(a) A \(3.00-\mu \mathrm{F}\) capacitor is connected to a \(12.0-\mathrm{V}\) battery. How much energy is stored in the capacitor? (b) If the capacitor had been connected to a \(6.00-\mathrm{V}\) battery, how much energy would have been stored?

Two identical parallel-plate capacitors, each with capacitance \(C,\) are charged to potential difference \(\Delta V\) and connected in parallel. Then the plate separation in one of the capacitors is doubled. (a) Find the total energy of the system of two capacitors before the plate separation is doubled. (b) Find the potential difference across each capacitor after the plate separation is doubled. (c) Find the total energy of the system after the plate separation is doubled. (d) Reconcile the difference in the answers to parts (a) and (c) with the law of conservation of energy.

An isolated charged conducting sphere of radius 12.0 \(\mathrm{cm}\) creates an electric field of \(4.90 \times 10^{4} \mathrm{N} / \mathrm{C}\) at a distance 21.0 \(\mathrm{cm}\) from its center. (a) What is its surface charge density? (b) What is its capacitance?

(a) Two spheres have radii \(a\) and \(b\) and their centers are a distance \(d\) apart. Show that the capacitance of this system is $$C=\frac{4 \pi \epsilon_{0}}{\frac{1}{a}+\frac{1}{b}-\frac{2}{d}}$$ provided that \(d\) is large compared with \(a\) and \(b\) . (Suggestion: Because the spheres are far apart, assume that the potential of each equals the sum of the potentials due to each sphere, and when calculating those potentials assume that \(V=k_{e} Q / r\) applies. (b) Show that as \(d\) approaches infinity the above result reduces to that of two spherical capacitors in series.

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.